009A Sample Final A, Problem 1

1. Find the following limits:
   (a)   $\lim _{x\rightarrow 0}{\frac {\tan(3x)}{x^{3}}}.$

   (b) $\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{6}+6x^{2}+2}}{x^{3}+x-1}}.$

   (c)   $\lim _{x\rightarrow 3}{\frac {x-3}{{\sqrt {x+1}}-2}}.$

   (d)   $\lim _{x\rightarrow 3}{\frac {x-1}{{\sqrt {x+1}}-1}}.$

   (e)  $\lim _{x\rightarrow \infty }{\frac {5x^{2}-2x+3}{1-3x^{2}}}.$

Foundations:
When evaluating limits of rational functions, the first idea to try is to simply plug in the limit. In addition to this, we must consider that as a limit,
${\frac {1}{\infty }}=0,$
and
$\lim _{x\rightarrow 0^{\pm }}{\frac {1}{x}}\,=\,\pm \infty .$
In the latter case, the sign matters. Unfortunately, most (but not all) exam questions require more work. Many of them will evaluate to an indeterminate form, or something of the form
${\frac {0}{0}}$   or   ${\frac {\pm \infty }{\pm \infty }}.$
In this case, here are several approaches to try:
We can multiply the numerator and denominator by the conjugate of the denominator. This frequently results in a term that cancels, allowing us to then just plug in our limit value. We can factor a term creatively. For example, $x-1$ can be factored as $\left({\sqrt {x}}-1\right)\left({\sqrt {x}}+1\right)$ , or as $\left({\sqrt[{3}]{x}}-1\right)\left(\left({\sqrt[{3}]{x}}\right)^{2}+{\sqrt[{3}]{x}}+1\right)$ , both of which could result in a factor that cancels in our fraction. We can apply l'Hôpital's Rule: Suppose $c$ is contained in some interval $I$. If $\lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0{\text{ or }}\pm \infty$  and $\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$   exists, and $g'(x)\neq 0$  for all $x\neq c$  in $I$, then $\lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$.
Note that the first requirement in l'Hôpital's Rule is that the fraction must be an indeterminate form. This should be shown in your answer for any exam question.

Part (a):
Note that both the numerator and denominator are continuous functions, and that the limit of each is 0 as $x$ approaches 0. This is an indeterminate form, and we can apply l'Hôpital's Rule:
$\lim _{x\rightarrow 0}{\frac {\tan(3x)}{x^{3}}}\,\,{\overset {l'H}{=}}\,\,\lim _{x\rightarrow 0}{\frac {\sec ^{2}(3x)\cdot 3}{3x^{2}}}\,=\,\lim _{x\rightarrow 0}{\frac {3}{3x^{2}}}.$
Now, $x^{2}$ can only be positive, so our limit can also only be positive. Thus, the limit is $+\infty$ .

Part (b):
In the case of limits at infinity, we can apply one other method. We can multiply our original argument by a fraction equal to one, and then can evaluate each term separately. Since we only need to consider values which are negative, we have that
$-\,{\frac {\sqrt {\frac {1}{x^{6}}}}{\frac {1}{x^{3}}}}\,=\,1,$
since for negative values of $x$,
${\sqrt {\frac {1}{x^{6}}}}\,=\,\left|{\frac {1}{x^{3}}}\right|\,=\,-\,{\frac {1}{x^{3}}}.$
This means that
$\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{6}+6x^{2}+2}}{x^{3}+x-1}}\,=\,\lim _{x\rightarrow -\infty }{\frac {\sqrt {x^{6}+6x^{2}+2}}{x^{3}+x-1}}\cdot \left(-\,{\frac {\sqrt {\frac {1}{x^{6}}}}{\frac {1}{x^{3}}}}\right)$
$=\,\lim _{x\rightarrow -\infty }-\,{\frac {\sqrt {1+{\frac {6x^{2}}{x^{6}}}+{\frac {2}{x^{6}}}}}{1+{\frac {x}{x^{3}}}+{\frac {1}{x}}}}$
$=\,\lim _{x\rightarrow -\infty }-\,{\frac {\sqrt {1+{\frac {6}{x^{4}}}+{\frac {2}{x^{6}}}}}{1+{\frac {1}{x^{2}}}+{\frac {1}{x}}}}$
$=\,-\,{\frac {\sqrt {1+0+0}}{1+0+0}}$
$=\,-1.$

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